a:x-1,5=6,5
=>x=6,5+1,5=8
b: \(2^3\cdot2^x-2,1=13,9\)
=>\(2^x\cdot8=13,9+2,1=16\)
=>\(2^x=\dfrac{16}{8}=2=2^1\)
=>x=1
c: \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
=>\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}=-\dfrac{28}{35}-\dfrac{15}{35}=-\dfrac{43}{35}\)
=>\(x=-\dfrac{43}{35}:\dfrac{7}{5}=-\dfrac{43}{35}\cdot\dfrac{5}{7}=\dfrac{-43}{49}\)
d: \(\dfrac{1}{9}\cdot3^4\cdot3^x=3^7\)
=>\(3^x\cdot\dfrac{3^4}{3^2}=3^7\)
=>\(3^x\cdot3^2=3^7\)
=>x+2=7
=>x=7-2=5
e: \(\left(\dfrac{6}{5}\right)^x=\dfrac{216}{125}\)
=>\(\left(\dfrac{6}{5}\right)^x=\left(\dfrac{6}{5}\right)^3\)
=>x=3
f: \(-\dfrac{2}{3}:x+\dfrac{5}{8}=-\dfrac{7}{12}\)
=>\(-\dfrac{2}{3}:x=-\dfrac{7}{12}-\dfrac{5}{8}=-\dfrac{14}{24}-\dfrac{15}{24}=-\dfrac{29}{24}\)
=>\(x=\dfrac{29}{24}:\dfrac{2}{3}=\dfrac{29}{24}\cdot\dfrac{3}{2}=\dfrac{87}{48}=\dfrac{29}{16}\)
